Q176 / 180 · COMEDK 2023 (Morning Shift) · Physics
In an electrical circuit $R, L, C$ and $\mathrm{AC}$ voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\pi / 3$. If instead $C$ is removed from the circuit, the phase difference is again $\pi / 3$. The power factor of the circuit is
- A$\dfrac{\sqrt{3}}{2}$
- B$1 / 2$
- Correct answerC1
- D$1 \sqrt{2}$
Solution
In an $LCR$ series circuit, the phase difference $\phi$ is given by $\tan \phi = \dfrac{X_L - X_C}{R}$.
When $L$ is removed, the circuit contains only $R$ and $C$. The phase difference is $\phi_1 = -\pi / 3$. Thus, $\tan(-\pi / 3) = \dfrac{-X_C}{R}$, which implies $\dfrac{X_C}{R} = \sqrt{3}$, so $X_C = R\sqrt{3}$.
When $C$ is removed, the circuit contains only $R$ and $L$. The phase difference is $\phi_2 = \pi / 3$. Thus, $\tan(\pi / 3) = \dfrac{X_L}{R}$, which implies $\dfrac{X_L}{R} = \sqrt{3}$, so $X_L = R\sqrt{3}$.
In the original $LCR$ series circuit, the net reactance is $X = X_L - X_C = R\sqrt{3} - R\sqrt{3} = 0$.
The circuit is in resonance, so the phase difference $\phi = 0$. The power factor is $\cos \phi = \cos 0 = 1$.
Answer: 1
When $L$ is removed, the circuit contains only $R$ and $C$. The phase difference is $\phi_1 = -\pi / 3$. Thus, $\tan(-\pi / 3) = \dfrac{-X_C}{R}$, which implies $\dfrac{X_C}{R} = \sqrt{3}$, so $X_C = R\sqrt{3}$.
When $C$ is removed, the circuit contains only $R$ and $L$. The phase difference is $\phi_2 = \pi / 3$. Thus, $\tan(\pi / 3) = \dfrac{X_L}{R}$, which implies $\dfrac{X_L}{R} = \sqrt{3}$, so $X_L = R\sqrt{3}$.
In the original $LCR$ series circuit, the net reactance is $X = X_L - X_C = R\sqrt{3} - R\sqrt{3} = 0$.
The circuit is in resonance, so the phase difference $\phi = 0$. The power factor is $\cos \phi = \cos 0 = 1$.
Answer: 1
Hence, the correct answer is option: 'C'.